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Edit: According to comment of Pace Nielsen, I remove question 2 of the previous version:

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0} M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}$ where $p_{0}=a,\;p_{n}=b$.

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

2. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?

**Note:**Inspired by methods from K theory, I tried to construct a functor $NK$ based on the constructions above. please see A functor on the category of rings, algebras or compact Hausdorff topological space

Perhaps, it would be interesting to ask "Is there a kind of periodicity property for this functor?"

Edit: According to comment of Pace Nielsen, I remove question 2 of the previous version:

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0} M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}$ where $p_{0}=a,\;p_{n}=b$.

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

2. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?

**Note:**Inspired by methods from K theory, I tried to construct a functor $NK$ based on the constructions above. please see A functor on the category of rings, algebras or compact Hausdorff topological space

Perhaps, it would be interesting to ask "Is there a kind of periodicity property for this functor?"

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0} M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}$ where $p_{0}=a,\;p_{n}=b$.

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

2. What is an example of a unital ring with a nilpotent element $a$ which is not equivalent to $0$?

3. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?

Edit: According to comment of Pace Nielsen, I remove question 2 of the previous version:

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0} M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}$ where $p_{0}=a,\;p_{n}=b$.

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

2. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?

**Note:**Inspired by methods from K theory, I tried to construct a functor $NK$ based on the constructions above. please see A functor on the category of rings, algebras or compact Hausdorff topological space

Perhaps, it would be interesting to ask "Is there a kind of periodicity property for this functor?"

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0}=a M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}=b$

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

2. What is an example of a unital ring with a nilpotent element $a$ which is not equivalent to $0$?

3. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?

Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:

We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of a ring). The relation $M$ is a reflexive and symmetric relation but is not a transitive relation. So we consider its saturation. The saturation of this relation is an equivalent relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$ with $p_{0} M p_{1},\;\;\;p_{1} M p_{2},\ldots p_{n-1} M p_{n}$ where $p_{0}=a,\;p_{n}=b$.

Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$

(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$ with $A=XY$ and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a Hilbert space. The same is true by replacing $B(H)$ with an arbitrary Von Neumann algebra. The same also is true without any topological consideration, that is by replacing $B(H)$ with $L(V)$, the space of linear endomorphisms of a vector space $V$.)

So for $R=M_{n}(\mathbb{C}),\;\;[0]$ is an algebraic variety,i.e: the variety of nilpotent matrices $A^{n}=0$

1.Is every equivalent class of $M_{n}(\mathbb{C})$ an algebraic variety?(the zero set of polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

2. What is an example of a unital ring with a nilpotent element $a$ which is not equivalent to $0$?

3. Assume that $A$ is a $C^{*}$ algebra and $a\in A$ satisfies $a^{k}=0$ for some $k>1$. Are there two elements $x,y \in A$ with $a=xy$ and $(yx)^{k-1}=0$?